Answer:
the first one is -4 2. is -2 3. is -70, 4. 20 5. 6 6. -45 7. 35 8. -20
Step-by-step explanation:
Answer:
(B)5200
Step-by-step explanation:
Since the value was increased by 30%, the new value in the options is 130% of the initial Value.
Let x be the number of hex nuts in the bin before that delivery.
130% of x = New Value
![\frac{130}{100} \: of\: x =New \: Value](https://tex.z-dn.net/?f=%5Cfrac%7B130%7D%7B100%7D%20%5C%3A%20of%5C%3A%20x%20%3DNew%20%5C%3A%20Value)
1.3 of x = New Value
x= New Value ÷ 1.3
Since the number of hex nuts in the bin is a whole number, we test the options to see which satisfies this condition.
- 5100 ÷ 1.3 =3923.08
- 5200 ÷ 1.3=4000
- 5400÷ 1.3=4153.85
- 5500÷ 1.3=4230.77
- 5600÷ 1.3=4307.69
5200 is the only option which satisfies our stated condition.
7 2 2
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3 x 1 is the answer to your question
Answer with Step-by-step explanation:
Let the ones digit of the number be y
and tens digit of the number be x.
The tens digit is two less than the units digit.
i.e. x=y-2
The value of the number exceeds twice the sum of its digits by 19.
i.e. 10x+y=2(x+y)+19
10(y-2)+y=2(y-2+y)+19
10y-20+y=2(2y-2)+19
11y-20=4y-4+19
11y-4y=20-4+19
7y=35
dividing both sides by 7, we get
y=5
Putting value of y in x=y-2, we get
x=3
10x+y=10×3+5
=35
Hence, the two digit number is:
35